Answer
See below
Work Step by Step
Define $T:R^5 \rightarrow V$ by $T(a,b,c,d,e)=ax^8+bx^6+cx^4+dx^2+e$
Assume $x_1=(a_1,b_1,c_1,d_1,e_1)\\
x_2=(a_2,b_2,c_2,d_2,e_2) \in R^5\\
\alpha \in R$
Obtain: $T(x_1+x_2)=T((a_1,b_1,c_1,d_1,e_1)+(a_2,b_2,c_2,d_2,e_2))\\
=T(a_1+a_2,b_1+b_2,c_1+c_2,d_1+d_2,e_1+e_2)\\
=(a_1+a_2)x^8+(b_1+b_2)x^6+(c_1+c_2)x^4+(d_1+d_2)x^2+(e_1+e_2)\\
=a_1x^8+b_1x^6+c_1x^4+d_1x^2+e_1+a_2x^8+b_2x^6+c_2x^4+d_2x^2+e_2\\
=T(a_1,b_1,c_1,d_1,e_1)+T(a_2,b_2,c_2,d_2,e_2)$
then $T(\alpha x_1)=T(\alpha (a_1,b_1,c_1,d_1,e_1)\\
=T(\alpha a_1+\alpha b_1+\alpha c_1+\alpha d_1+\alpha e_1)\\
=\alpha a_1x^8+\alpha b_1x^6+\alpha c_1x^4+\alpha d_1x^2+\alpha e_1\\
=\alpha (a_1x^8+b_1x^6+c_1x^4+d_1x^2+e_1)\\
=\alpha T(a_1,b_1,c_1,d_1,e_1)\\
=\alpha T(x_1)$
Hence, $T$ is a linear transformation.
Obtain: $Ker(T)=\{x \in R^5:T (x)=0\}\\
=\{(a,b,c,d,e) \in R^5: T(a,b,c,d,e)=0\}\\
=\{(a,b,c,d,e) \in R^5: ax^8+bx^6+cx^4+dx^2+e\\
\rightarrow a=b=c=d=e=0$
then $Ker(T)=\{(0,0,0,0,0)\}=\{0\}$
$T$ is one-to-one (1)
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^5\\
0+\dim [Rng(T)]=5\\
\dim [Rng(T)]=5$
$Rng(T) \subset V$ and $\dim R^5=5=\dim [Rng(T)]$
$T$ is onto (2)
From $(1),(2)$, $T$ is an isomorphism.
